NOTES ON A FAMILY OF INHOMOGENEOUS LINEAR ORDINARY DIFFERENTIAL EQUATIONS
FENG QI
Institute of Mathematics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China; College of Mathematics, Inner Mongolia University for Nationalities, Tongliao City, Inner Mongolia Autonomous Region, 028043, China;
Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin City, 300387, China
JING-LIN WANG
Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin City, 300387, China
BAI-NI GUO
School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China
Abstract. In the note, the authors find several simple and explicit forms for a family of inhomogeneous linear ordinary differential equations studied in “D. Lim,Differential equations for Daehee polynomials and their applications, J. Nonlinear Sci. Appl. 10 (2017), no. 4, 1303–1315; Available online at
http://dx.doi.org/10.22436/jnsa.010.04.02”.
1. Motivation and main results
In [3, Theorem 1], it was obtained inductively and recurrently that the differential equations
∂nF(t, x) ∂ tn =
1 tn
" n X i=0
ai(n, x) +
bi(n, x) ln(1 +t)
t
1 +t i#
F(t, x), n≥0 (1.1)
E-mail addresses:qifeng618@gmail.com, qifeng618@hotmail.com, qifeng618@qq.com, jing-lin.wang@hotmail.com, bai.ni.guo@gmail.com, bai.ni.guo@hotmail.com. 2010Mathematics Subject Classification. 11B73, 11M35, 26A48, 26D07, 34A05, 44A10. Key words and phrases. simple form; explicit form; differential equation; Lerch transcendent; logarithmic function.
have the same solution
F(t, x) =ln(1 +t) t (1 +t)
x, (1.2)
wherea0(n, x) = (−1)nn! and b0(n, x) = 0 for alln≥0,
aj(n, x) = (x)j(−1)n−j(n−j)! n−j X ij−1=0
n−j−ij−1
X ij−2=0
· · ·
n−j−ij−1−···−i2
X ii=0
(n−ij−1− · · · −i1−j+ 1)
and
bj(n, x) = j−1 X k=0
j−1 Y i=0
(x−i)(−1)n−j(n−j)! x−k
n−j X ij−1=0
n−j−ii−1
X ij−2=0
· · ·
n−j−ij−1−···−i2
X i1=0
(n−ij−1− · · · −i1−j+ 1)
for 1≤j≤n+ 1, and (x)n =Q n−1
k=0(x−k) is the falling factorial.
It is clear that the above expressions foraj(n, x) and bj(n, x) are very difficult to compute by hand or by computer. The derivation of the quantitiesaj(n, x) and bj(n, x) in [3] is much long and tedious.
The aim of this paper is to alternatively supply several new, simple, and explicit expressions for aj(n, x) and bj(n, x). In other words, the aim of this paper is to alternatively provide several new, simple, and explicit forms for the family of differential equations in (1.1).
Our main results can be stated as the following theorems.
Theorem 1. Forn≥0, the function F(t, x)defined in (1.2)satisfies ∂nF
∂ tn = (−1) nn!
tn ( n
X i=0
(−1)i(x)i i!
"
1− 1
ln(1 +t) n−i X k=1 1 k
t
1 +t
k# t
1 +t i)
F (1.3)
and
∂nF ∂ tn =
n! (1 +t)n
( n X i=0
(−1)i (x)n−i (n−i)!
"
1− 1
ln(1 +t) i X k=1 1 k
t
1 +t
k#1 +t t
i) F.
Theorem 2. Let
Φ(z, s, a) =
∞
X k=0
zk
(a+k)s, a6= 0,−1, . . .
denote the Lerch transcendent. Then
∂nF(t, x) ∂ tn =
(−1)nn! (1 +t)n−x+1
n X i=0
(−1)i(x)i i! Φ
t
1 +t,1, n−i+ 1
and
∂nF(t, x) ∂ tn =
n! (1 +t)n−x+1
n X
i=0
(−1)i (x)n−i (n−i)!Φ
t
1 +t,1, i+ 1
2. A lemma
In order to prove our main results in Theorems 1 and 2, we need the following lemma.
Lemma 1([7, Theorem 2]). Let
f(x) =
lnx
x−1, 0< x6= 1; 1, x= 1.
Forn≥0, the nth derivative off(x)can be computed by
f(n)(x) =
(−1)nn! (x−1)n+1
"
lnx− n X k=1 1 k
x−1 x
k#
, 0< x6= 1;
(−1)n n!
n+ 1, x= 1
(2.1)
and
f(n)(x) =
(−1)n n! xn+1Φ
x−1
x ,1, n+ 1
, 0< x6= 1;
(−1)n n!
n+ 1, x= 1.
(2.2)
3. Proofs of Theorems 1 and 2
We now start out to prove our main results in Theorems 1 and 2.
Proof of Theorem 1. By (2.1), it is straightforward that
∂nF(t, x) ∂ tn =
n X i=0
n i
ln(1 +t) t
(n−i)
[(1 +t)x](i)
= n X i=0 n i
(−1)n−i(n−i)! tn−i+1
"
ln(1 +t)− n−i X k=1 1 k t 1 +t
k#
(x)i(1 +t)x−i
= (1 +t)
xln(1 +t)
t
(−1)nn! tn
n X
i=0
(−1)i(x) i i!
"
1− 1
ln(1 +t) n−i X k=1 1 k
t
1 +t
k# t
1 +t i
= (−1)nn! tn
( n X i=0
(−1)i(x)i i!
"
1− 1
ln(1 +t) n−i X k=1 1 k
t
1 +t
k# t
1 +t i)
F(t, x)
and
∂nF(t, x) ∂ tn =
n X i=0 n i
ln(1 +t) t
(i)
[(1 +t)x](n−i)
= n X i=0 n i
(−1)ii! ti+1
"
ln(1 +t)− i X k=1 1 k t
1 +t k#
(x)n−i(1 +t)x−n+i
= ln(1 +t) t(1 +t)n−x
n X i=0
n i
(−1)ii! ti
"
1− 1
ln(1 +t) i X k=1 1 k t
1 +t k#
(x)n−i(1 +t)i
= F(t, x) (1 +t)n
n X i=0
(−1)i n! (n−i)!
"
1− 1
ln(1 +t) i X k=1 1 k t
1 +t k#
(x)n−i 1 +t
= n! (1 +t)n
( n X
i=0
(−1)i (x)n−i (n−i)!
"
1− 1
ln(1 +t) i X k=1 1 k
t
1 +t
k#1 +t t
i) F(t, x).
The proof of Theorem 1 is complete.
Proof of Theorem 2. By (2.2), it is immediate that
∂nF(t, x) ∂ tn =
n X i=0
n i
ln(1 +t) t
(n−i)
[(1 +t)x](i)
= n X i=0
n i
(−1)n−i (n−i)! (1 +t)n−i+1Φ
t
1 +t,1, n−i+ 1
(x)i(1 +t)x−i
= (−1) nn!
(1 +t)n−x+1 n X i=0
(−1)i(x)i i! Φ
t
1 +t,1, n−i+ 1
and
∂nF(t, x) ∂ tn =
n X i=0
n
i
ln(1 +t) t
(i)
[(1 +t)x](n−i)
= n X i=0
n i
(−1)i i! (1 +t)i+1Φ
t
1 +t,1, i+ 1
(x)n−i(1 +t)x−n+i
= n!
(1 +t)n−x+1 n X i=0
(−1)i (x)n−i (n−i)!Φ
t
1 +t,1, i+ 1
.
The proof of Theorem 2 is complete.
4. Remarks
Finally we give several remarks about our main results mentioned and verified above.
Remark 1. Comparing (1.1) with (1.3) reveals that the formulas
ai(n, x) = (−1)n−i n!
i!(x)i and
bi(n, x) =−ai(n, x) n−i X k=1 1 k
t
1 +t k
= (−1)n−i+1n! i!(x)i
n−i X k=1 1 k
t
1 +t k
should be valid for alli, n≥0.
Remark 2. The idea of this paper comes from the articles [1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14] and the closely related references therein.
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c
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